who know how to convert z/(z^2+9) to z/(z+3i)/(z-3i)?i really do not have any idea how the book convert it.
Woah, it's better when you put the parenthesis where it has to go
$\displaystyle \frac{1}{z^2+1}=\frac{1}{(z-i)(z+i)}=\frac{a+bi}{z-i}+\frac{c+di}{z+i}$
$\displaystyle =\frac{(a+bi)(z+i)+(c+di)(z-i)}{z^2+1}$
The numerator has to be equal to 1.
So develop, then group the terms including z and write that its coefficient must be null since z doesn't appear in "1".
Solve for a, b, c and d